Precalculus Examples

$\ln (x + 1) + \ln (x - 1) = \ln (8)$

Step 1

Simplify the left side.

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Step 1.1

Use the product property of logarithms, $\log_{b} (x) + \log_{b} (y) = \log_{b} (x y)$ .

$\ln ((x + 1) (x - 1)) = \ln (8)$

Step 1.2

Expand $(x + 1) (x - 1)$ using the FOIL Method.

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Step 1.2.1

Apply the distributive property.

$\ln (x (x - 1) + 1 (x - 1)) = \ln (8)$

Step 1.2.2

Apply the distributive property.

$\ln (x \cdot x + x \cdot - 1 + 1 (x - 1)) = \ln (8)$

Step 1.2.3

Apply the distributive property.

$\ln (x \cdot x + x \cdot - 1 + 1 x + 1 \cdot - 1) = \ln (8)$

Step 1.3

Simplify and combine like terms.

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Step 1.3.1

Simplify each term.

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Step 1.3.1.1

Multiply $x$ by $x$ .

$\ln (x^{2} + x \cdot - 1 + 1 x + 1 \cdot - 1) = \ln (8)$

Step 1.3.1.2

Move $- 1$ to the left of $x$ .

$\ln (x^{2} - 1 \cdot x + 1 x + 1 \cdot - 1) = \ln (8)$

Step 1.3.1.3

Rewrite $- 1 x$ as $- x$ .

$\ln (x^{2} - x + 1 x + 1 \cdot - 1) = \ln (8)$

Step 1.3.1.4

Multiply $x$ by $1$ .

$\ln (x^{2} - x + x + 1 \cdot - 1) = \ln (8)$

Step 1.3.1.5

Multiply $- 1$ by $1$ .

$\ln (x^{2} - x + x - 1) = \ln (8)$

Step 1.3.2

Add $- x$ and $x$ .

$\ln (x^{2} + 0 - 1) = \ln (8)$

Step 1.3.3

Add $x^{2}$ and $0$ .

$\ln (x^{2} - 1) = \ln (8)$

Step 2

For the equation to be equal, the argument of the logarithms on both sides of the equation must be equal.

$x^{2} - 1 = 8$

Step 3

Solve for $x$ .

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Step 3.1

Move all terms not containing $x$ to the right side of the equation.

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Step 3.1.1

Add $1$ to both sides of the equation.

$x^{2} = 8 + 1$

Step 3.1.2

Add $8$ and $1$ .

$x^{2} = 9$

Step 3.2

Take the specified root of both sides of the equation to eliminate the exponent on the left side.

$x = \pm \sqrt{9}$

Step 3.3

Simplify $\pm \sqrt{9}$ .

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Step 3.3.1

Rewrite $9$ as $3^{2}$ .

$x = \pm \sqrt{3^{2}}$

Step 3.3.2

Pull terms out from under the radical, assuming positive real numbers.

$x = \pm 3$

Step 3.4

The complete solution is the result of both the positive and negative portions of the solution.

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Step 3.4.1

First, use the positive value of the $\pm$ to find the first solution.

$x = 3$

Step 3.4.2

Next, use the negative value of the $\pm$ to find the second solution.

$x = - 3$

Step 3.4.3

The complete solution is the result of both the positive and negative portions of the solution.

$x = 3, - 3$

Step 4

Exclude the solutions that do not make $\ln (x + 1) + \ln (x - 1) = \ln (8)$ true.

$x = 3$